[S.T. 4 ] - Capacitive and Inductive filtered rectifiers

Discussion in 'Homework Help' started by PsySc0rpi0n, Mar 16, 2015.

  1. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    My next task is to analyse this kind of circuits...

    First question is:

    Is it possible to measure the minimum value of v_out waveform using ".meas" command?

    I know how to do it for MAX values:

    Code (Text):
    1. .meas v_out MAX V(out)
    And what about for minimum value?

    Edited;
    What I need is to measure MAX value and MIN value of a rectified waveform and check the V ripple. With these values I could easily find V ripple!
     
    Last edited: Mar 16, 2015
  2. ericgibbs

    AAC Fanatic!

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    hi psy,
    Please post the LTS asc file for the circuit you are testing.
    E
     
  3. PsySc0rpi0n

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    Here it is!
     
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  4. ericgibbs

    AAC Fanatic!

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    hi,
    After you have run the sim, use Menu View/Spice Error log to see your measure results.
     
  5. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I know about error log to see measures but v_ripple_min cannot be zero, right? That's why I think "MIN" directive is not being processed by LTSpice as "MAX" is!

    But the values that I have in the screen attached are with cursors manually positioned in the plot. So, that might be not as accurate as measure command would be!
     
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    Last edited: Mar 17, 2015
  6. ericgibbs

    AAC Fanatic!

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    You have included the zero start voltage in your .Tran!!!

    Change your .tran to say 60m , start 20m, do not include the zero start period.
     
    PsySc0rpi0n likes this.
  7. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, I got it...

    Now I need to know how to calculate the cap value so that Vripple is about 0% of it's max value...

    As far I as understand, Vripple our teacher wants is:

    Vripple.gif

    Now how do I calculate C, so that Vripple = 0.684V???
     
  8. ericgibbs

    AAC Fanatic!

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    You could use the formula Q = C *V, where Q= I * t, this assumes a linear discharge from Vmax to Vmin.

    So work out the average current over the period between Vmax and Vmin and use the above formula.
    Post what you the think the Cap value is and to confirm that it is close enough change the Cap value in the Sim and re-run the sim.

    Extend the .tran period to get a more accurate value for I_load and Vmax/min, say 200mS
     
    Last edited: Mar 17, 2015
  9. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Well, I did it like this:

    I_load.gif

    Then:

    cap_value.gif

    But when I change the cap value to 343.7uF, the vripple is about 1.12V and not 0.684V as previously calculated! What is wrong?
     
  10. ericgibbs

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    The Iavg is close enough, but where did you get that second formula.?

    Retry using C= [I * t]/Vrip, where t is ~ 1/5o
     
  11. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    2nd formula is here in Vripple section
    http://www.zen22142.zen.co.uk/Design/dcpsu.htm

    and this link was taken from some thread from AAC that google returned to me!

    Ok, let me try using that formula!


    Edited;
    Ok, that formula gives me correct results, but what's wrong with the other one?


    Edited2;

    Ok, I think that the formula I used is specific for full-wave rectifiers... I'm using half-wave rectifier, so that "2" in the formula I used was messing my calcs up!

    Please, confirm this with me, Mr. EricGibbs!

    Close enough is perfect, as your signature says!
     
    Last edited: Mar 18, 2015
  12. ericgibbs

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    I use this formula for Vripple Cap value.

    Time = (R * C) * ln(Vmin/Vmax)

    Transpose for Cap and compare the result from your corrected formula, lets see how they compare.:)

    In practice most high value electrolytic caps have a tolerance values of +/-20% or higher.!
    http://www.electronics-tutorials.ws/capacitor/cap_3.html
     
    Last edited: Mar 17, 2015
  13. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    In is the current across the load resistor?
    Vmin and Vmax are the V outs as if the circuit has no capacitor????
    And Time is what??? Time between peaks???
     
  14. ericgibbs

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    Its not In its Ln [ natural logarithm]:)

    Vmax and Vmin are the smoothed voltage on the capacitor, example Vmax = 7.7V and Vmin = 7V , Natural Log, Ln(7/7.7)
    Vripple = Vmax -Vmin

    If you recall from a previous question, for a 50Hz frequency the Period is 20mSec, but using a half wave rectifier the forward conduction period is 'ideally' only 10mSec, but due to the Vfwd of the diode this is reduced.

    The other factor to consider is the 're-charging' period of the capacitor, during this time the positive cycle is supplying current to recharge the capacitor and the load [330R]. Using LTS to measure this gives ~ 2.4mS recharge time.

    If you now measure from the Vmax to the Vmin of the plot you will see that the actual capacitor discharge time is ~17.7mSec, this the time for the equation.

    Note the SUM of recharge time and discharge times is 20mSec
     
  15. PsySc0rpi0n

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    Mar 4, 2014
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    Apart of all those details, I'm just looking for an approximate value for the capacitor that gives me a ripple of 10% of it's max value!!! I think I got that using the above formulas you suggested. But If I use a full wave rectifier, the formula for C is a bit different. It has the f_out difference which is double. In my case is 20mS. In full-wave is 10mS...

    Note:

    20mS period are 50Hz, right?
    Doubling that period, I'll get 10mS, right?

    So, how many Hertz is an half period of 20mS???

    If 20mS are 50Hz
    Then 10mS are x
    x = (50*10)/20 = 25Hz...
    Not good. Should be 100Hz...

    Where the hell am I going wrong here??? I'm going mad!
     
  16. ericgibbs

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    hi Psy,
    As you know, the Period of a Sine wave is the inverse of the Frequency, ie: 1/50 = 0.02Secs

    With a single diode rectifier driven by a 50Hz voltage source, only conducts on the positive half cycle, so the Period between capacitor charging pulses is 20mSec. [10mSec forward biased, 10mSec reverse biased]

    With a bridge/dual diode rectifier the diode/s conduct on both half cycles of the input Sine wave, so the Period between the capacitor charging pulses is 10mSec.

    You are not going mad.. you are over thinking the problem.;)

    Eric

    EDIT:
    This image may help.
    Half wave and Full wave rectification.
    I have used a 600uF cap on both circuits.
     
    Last edited: Mar 18, 2015
  17. PsySc0rpi0n

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    Mar 4, 2014
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    I understand the theory which makes all sense... What I'm not understanding and what is leaving me mad is the math I did in the previous post...

    I should get 100Hz for 10mS but I'm getting 25Hz instead!
     
  18. ericgibbs

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    Period= 1/Frequency

    Per1 = 1/100 = 0.01 Sec
    Per2 = 1/25 = 0.04 Sec

     
  19. PsySc0rpi0n

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    Mar 4, 2014
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    Why you say it's (50*0.02)/0.01???

    20mS ↔ 50Hz
    10mS ↔ xHz

    20×x = 50×10⇔x=500/20⇔x=25
     
  20. ericgibbs

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    Because a 50Hz sine wave has a Period of 20mSec, in 1 second there will be 50 periods of 20mSec.

    So there must be 100 periods of 10mSec in 1 second

    is incorrect.!
    You are confusing complete cycles of a 50Hz SINE wave, with a Period of 20mSec with 100 rectified half cycles of 10mSec
     
    Last edited: Mar 18, 2015
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